※ 引述《gn00648013 (大偉)》之銘言:
: 題目是
: ----------------------------------------
: x'=-2x+y
: y'=-4x+3y+e^-t x'=dx/dt y'=dy/dt
: ----------------------------------------
改寫成
x'+2x-y=0
y'+4x-3y=e^-t
(D+2)x-y=0
=> 4x+(D-3)y=e^-t
│(D+2) -1 │ │0 -1 │
│4 (D-3)│x = │e^-t (D-3) │
=>(D^2-D-2)x=e^-t
Xh=C1e^2t+C2e^-t
Xp= [1/(D-2)(D+1)]e^-t
=(-1/3)xe^-t
X=C1e^2t+C2e^-t +(-1/3)xe^-t
Y同樣的方法解
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